After doing that I get: the integral from 0 to 1 of [(-2/pi) sqr(1 + u^2pi) du]. Would I have to do a trig substitution?
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KrystenMar 7 '11 at 20:35

I don't think you should have a $\pi$ under the square root, and $\sqrt{1+u^2}du$ may be in your tables, or you can substitute $u=\tan(\theta)$ But I think the limits in $u$ are not 0 to 1.
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Ross MillikanMar 7 '11 at 20:49